Gracy3535:

S=1/2gt^2
S=1/2*10*2.2^2
S=5*4.84
S=24.2

...great sis....as simple as it is,I missed it...why? 2.2^ 2 for my mind is 8...dat's too know...

Damayogenius18:

7. A physics student standing on the edge of a cliff throws a stone vertically downward with an
initial speed of 10.0 m/s. The instant before the stone hits the ground below, it is traveling at
a speed of 30.0 m/s. If the physics student were to throw the rock horizontally outward from
the cliff instead, with the same initial speed of 10.0 m/s, what is the magnitude of the velocity
of the stone just before it hits the ground...A..10..B..20..C..30..D..40

Vy = final velocity of the vertical component of the velocity.

Uy = initial velocity of the vertical component

Ux = initial velocity of the horizontal component

Vx = final velocity of the horizontal component

when the body was thrown vertically downwards

(Vy)² = (Uy)² + 2gh.

30² = 10² + 2(10)h

20h = 800

h = 40m ( the height of the building )

Now, when the ball was thrown horizontally..

(Vy)² = (Uy)² + 2gh.

but Uy = 0 , since the ball was thrown horizontally.

(Vy)² = 2(10)(40)

(Vy)² = 800

Vy = √(800)ms-¹ (that is the final vertical component of the velocity )

But Ux = Vx (horizontal component is not affected by g)

The velocity it hits the ground V = √[(Vy)² + (Vx)²]

V = √[(√800)² + (10)²]

V = √(800 + 100)

V = √(900)

V = 30ms-¹

Damayogenius18:
...I don't get what u r implying boss...one aspect is about surface tension which I try to point out to you that it has formula..the other aspect is Tension which I said it also has multiple formula based on the topic...in sound,there is tension formula e.g string..f= √TL/ M..we can make T subject of formula to get tension...in uptrust, tension T= mg- pvg...in circular motion..T=mv^ 2/ r + mg... In a lift..T= m(a+g)...so on...so I don't understand what you are trying to say..

I never said surface tension has no formula, I was talking about "tension" which has no particular formula.

Damayogenius18:
...great sis....as simple as it is,I missed it...why? 2.2^ 2 for my mind is 8...dat's too know...

I understand you that happens alot. Now that one has to study without calculator

samuelizz:





Vy = final velocity of the vertical component of the velocity.

Uy = initial velocity of the vertical component

Ux = initial velocity of the horizontal component

Vx = final velocity of the horizontal component

when the body was thrown vertically downwards

(Vy)² = (Uy)² + 2gh.

30² = 10² + 2(10)h

20h = 800

h = 40m ( the height of the building )

Now, when the ball was thrown horizontally..

(Vy)² = (Uy)² + 2gh.

but Uy = 0 , since the ball was thrown horizontally.

(Vy)² = 2(10)(40)

(Vy)² = 800

Vy = √(800)ms-¹ (that is the final vertical component of the velocity )

But Ux = Vx (horizontal component is not affected by g)

The velocity it hits the ground V = √[(Vy)² + (Vx)²]

V = √[(√800)² + (10)²]

V = √(800 + 100)

V = √(900)

V = 30ms-¹

....boss u lost me o..I know resultant velocity v= √(vx^2 + vy^2)..do we need to calculate Vy again...I think it has been given as 30m/s..I think they ask the horizontal velocity the ball hits the ground not the resultant..

samuelizz:




Vy = final velocity of the vertical component of the velocity.
Uy = initial velocity of the vertical component
Ux = initial velocity of the horizontal component
Vx = final velocity of the horizontal component
when the body was thrown vertically downwards
(Vy)² = (Uy)² + 2gh.
30² = 10² + 2(10)h
20h = 800
h = 40m ( the height of the building )
Now, when the ball was thrown horizontally..
(Vy)² = (Uy)² + 2gh.
but Uy = 0 , since the ball was thrown horizontally.
(Vy)² = 2(10)(40)
(Vy)² = 800
Vy = √(800)ms-¹ (that is the final vertical component of the velocity )
But Ux = Vx (horizontal component is not affected by g)
The velocity it hits the ground V = √[(Vy)² + (Vx)²]
V = √[(√800)² + (10)²]
V = √(800 + 100)
V = √(900)
V = 30ms-¹

Thanx boss for the working. I actually did cross multiplcation to get 30m/s

samuelizz:


I never said surface tension has no formula, I was talking about "tension" which has no particular formula.

..OK boss,m sori I did nt understand u...

Gracy3535:

Thanx boss for the working. I actually did cross multiplcation to get 30m/s

..cross multiply as how..can you show ur working sis

Damayogenius18:
....boss u lost me o..I know resultant velocity v= √(vx^2 + vy^2)..do we need to calculate Vy again...I think it has been given as 30m/s..I think they ask the horizontal velocity the ball hits the ground not the resultant..

Vy is 30ms-¹ when it was thrown vertically downwards.

The new Vy I solved for was when it was thrown horizontally.

Gracy3535:

Thanx boss for the working. I actually did cross multiplcation to get 30m/s

LOL.
you're welcome.

samuelizz:


Vy is 30ms-¹ when it was thrown vertically downwards.

The new Vy I solved for was when it was thrown horizontally.

OK boss...y is vx now 10...are we also ask to calculate resultant velocity...the answer that was chosen is 30m/s..maybe I don't understand the question sha!

Damayogenius18:
..cross multiply as how..can you show ur working sis

pls dnt laugh @ me.
Uv=10, Vv=30, Uh=10 Vh=?
Uv=Vv
Uh=Vh
10=30
10=x
then i cross multiply
10x=300
x=300/10
x=30m/s
.'. Vh=30m/s
that is the ojoro i did.

Damayogenius18:
OK boss...y is vx now 10...are we also ask to calculate resultant velocity...the answer that was chosen is 30m/s..maybe I don't understand the question sha!

(Vx)² = (Ux)² + 2gh.

but g is always zero when it comes to horizontal components

(Vx)² = (Ux)² + 2(0)h

(Vx)² = (Ux)²

Vx = Ux.

samuelizz:


(Vx)² = (Ux)² + 2gh.

but g is always zero when it comes to horizontal components

(Vx)² = (Ux)² + 2(0)h

(Vx)² = (Ux)²

Vx = Ux.

Nw i get

samuelizz:


(Vx)² = (Ux)² + 2gh.

but g is always zero when it comes to horizontal components

(Vx)² = (Ux)² + 2(0)h

(Vx)² = (Ux)²

Vx = Ux.

..thanks boss..I pray am not like this o in exam hall..glad m making all the mistake now...m overtaxing my brain...we are told to calculate the magnitude of the velocity I.e resultant..I tot we are asked horizontal velocity when its abt to fall....I also forget the fact that horizontal velocity is always constant during projectile motion...

Gracy3535:
pls dnt laugh @ me.
Uv=10, Vv=30, Uh=10 Vv=?
Uv=Vv
Uh=Vv
10=30
10=x
then i cross multiply
10x=300
x=300/10
x=30m/s
.'. Vh=30m/s
that is the ojoro i did.

..of course,I do...u r very smart..I wld av left d question unanswered and u would av gotten...that's y is good to b smart...

Damayogenius18:
..of course,I do...u r very smart..I wld av left d question unanswered and u would av gotten...that's y is good to b smart...

Me that is learning from you i just did wuru wuru to the answer according to my teacher but tanx sha

samuelizz:

You can use lamlad, it is a very good text.

Thanks boss.

Damayogenius18:
...great sis....as simple as it is,I missed it...why? 2.2^ 2 for my mind is 8...dat's too know...

lol

Gracy3535:

Me that is learning from you i just did wuru wuru to the answer according to my teacher but tanx sha

...your teacher tot u wuru wuru method to get right option....pls can u recommend that teacher to me.....lol...

Damayogenius18:
...your teacher tot u wuru wuru method to get right option....pls can u recommend that teacher to me.....lol...

funny u. Actually what i meant was that if any of his student got an answer to his question with a wrong method/solution he usual called that type of job "wuru wuru to the answer".

@SPORE

You're sighted bro.
How's been the day?

Eben331:
lol

..bros,my brain is multi tasking 2day...both ques are simple BT my analysis are bad...m just so glad today is not post utme...

Ajet001:

@SPORE

You're sighted bro.
How's been the day?

...doc...any advice?

samuelizz:


(Vx)² = (Ux)² + 2gh.

but g is always zero when it comes to horizontal components

(Vx)² = (Ux)² + 2(0)h

(Vx)² = (Ux)²

Vx = Ux...

...thanks boss..are you a medical student or engineering?

Gracy3535:
funny u. Actually what i meant was that if any of his student got an answer to his question with a wrong method/solution he usual called that type of job "wuru wuru to the answer".

...cant u see?can decipher simple post...guess I need to leave nairaland now....one of those day..

samuelizz:


You can use lamlad, it is a very good text.

i cant find lamlad here. Pls is New school ok. Cus i dnt want to waste another money like that of essential

Eben331:
Boss ke. I am your boy o. From V2=V1(1+gamma*change in tita) V2/V1=1+gamma*change in tita---eqn (i) Know dat volume is inversely proportional to density, we can say dat d=M/V where M is mass which is a constant. M=dV d1V1=d2V2 V2/V1=d1/d2 Substitute for eqn(i): 1+gamma*change in tita=d1/d2 d1=d2(1+gamma*change in tita) You can use d1 to rep density of mercury at 0degree celsius and d2 to rep density of mercury at 100 degrees celsius and at 22 degrees celsius. SOLUTION: By making d2 the subject of formula: d2=d1/1+gamma+change in tita d2 at 100^oC=1.36*10^2/1+180*10^-6*100=136/1.018=1.33*10^2kg/m3 d2 at 22^oC=1.36*10^2/1+180*10^-6*22=136/1.00396=1.35*10^2kg/m3.

thanks boss

Damayogenius18:
...cant u see?can decipher simple post...guess I need to leave nairaland now....one of those day..

why I suggest u shuld let ur brain relax for 2-3 days then check out the result.

Gracy3535:
i cant find lamlad here. Pls is New school ok. Cus i dnt want to waste another money like that of essential

New school is okay too.
solve a lot of past questions

Damayogenius18:
...thanks boss..are you a medical student or engineering?

Let's just say engineering.

*PROJECTILES.*

I'm sure we are all familiar with the topic projectile.. But let's assume you don't know, now , what is projectile ?


Well, *projectile is the launching of a body into space which is under the influence of gravity.*

From that definition, it can be said that , when dealing with the topic projectile, we are going to be dealing with *g* (acceleration due to gravity) a lot.

Projectile is a 2 - dimensional motion, that is , it consists of the x , y components... Are you confused??... Just be calm, you will understand soon.

The x components is always the horizontal components , in a better understanding, the component parallel to the ground while the y components is always the vertical component i.e it is always perpendicular to the ground. ...


y.
| /
| /
| /
| /
|/___________y

Now, you have an idea of what I've been saying.

Let's say a body was projected with a velocity *U* and it makes an angle *B* with the horizontal.

When you resolve the velocity into the vertical and horizontal components using your trigonometry knowledge.

You will have .

Ux = UcosB (Ux is the horizontal component)

Uy = UsinB (Uy is the vertical component)

Right?.. Good, now let's proceed.

*TERMS USED IN PROJECTILE*

In every topic in physics, there are always some terms always used in the topics, well, projectile is not an exception!

1. *MAXIMUM HEIGHT* : From the word "maximum" and the word "height" I'm sure you are getting the definition already..

It can be defined as the *maximum vertical distance* which a body attains from the *point of projection*

Simple, right?

You should know that at maximum height, final velocity V is always zero because the body that was projected will stop momentarily before coming down!!

Let's now derive the formula for maximum height..

From our popular formula of motion, one of them is

v² = u² ± 2gh.
But since the body is projected upwards, that makes our *g* to be negative.

v² = u² - 2gh.

We also know that, when a body is projected upwards, and since the height is a vertical distance, it will be affected by *g*, only the horizontal distance is not affected by *g*, we will get to that.

We know that Uy = UsinB. And that max height Vy = 0

0² = (UsinB)² - 2gh

2gh = u²sin²B

h = u²sin²B/2g.

And that's the formula for maximum height.

2) *TIME OF FLIGHT*: It is the time required for the body to reach the same level of point of projection i.e the time taken for the journey.

In our motion formula, we have,

h = ut - ½gt²

At the end of the journey, the body will touch the ground again right?, do you agree with me?.. Good

If it touched the ground again, the vertical height with be 0 right?... Agreed?

Since we are making use of vertical motion, Uy = UsinB

0 = UsinB•t - ½gt²

UsinB = ½gt

t = 2usinB/g

And that is our time of flight.

NB : Time of flight is always twice the time it took to reach the maximum height.

3) *RANGE*: Range can be defined as the *maximum horizontal distance* from the point of projection in which the body projected touch the line of point at the *same horizontal level*

Please, it must always be at the *same level at which it was projected from*

Now let's derive the formula, we are still talking of projectile here..

From one of our motion formula.

We have ,

h = Uy•t - ½gt²

But since Range is represented with R.

And since it is an horizontal motion, g = 0, we substitute Uy as Ux

R = Ux•t - ½(0)t²

R = Ux•t

Where t is our time of flight, t = 2usinB/g and Ux = ucosB.

R= ucosB × 2usinB/g

R = 2u²sinBcosB/g

But 2sinBcosB = sin2B

R = u²sin2B/g

And that is our formula for Range..

Now, let me clarify something..
In a question , you might be asked for the horizontal distance covered and in another question, you will be asked for range.

Please , note that Horizontal distance is deferent from Range.

When it comes to Range, the body must always be at the *same level* with the point the body was projected..

But when it comes to *horizontal distance*, it not necessary that the body must be at the same level with the point at which the body was projected..

The formula for horizontal distance is R = Ux•t. I.e horizontal component of the velocity × the total time.

Lets take an example, let's say a body was projected upwards from the *top of a building* and it hits the *ground after some time. From that, you can see that the ball did not land at the same level with the point at which the body was projected...

So, they are different!!.